Steven D'Aprano wrote
You should always quote enough of the previous poster's message to give
context. Messages sometimes go missing in Usenet, and people won't have
the foggiest idea what you are talking about.
one would think that given how many Pythoneers we now have working
for google, at
On Thu, 09 Mar 2006 22:01:17 -0800, John wrote:
Thanks a lot,
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
I don't know. What is it?
You should always quote enough of the previous poster's message to give
context. Messages
I have two strings S1 and S2. I want to know how many times
S2 occurs inside S1.
For instance
if S1 =
and S2 = AA
then the count is 3. Is there an easy way to do this in python?
I was trying to use the count function but it does not do
overlapping counts it seems.
Thanks,
--j
--
John [EMAIL PROTECTED] writes:
if S1 =
and S2 = AA
then the count is 3. Is there an easy way to do this in python?
I was trying to use the count function but it does not do
overlapping counts it seems.
len([1 for i in xrange(len(s1)) if s1[i:].startswith(s2)])
--
Thanks a lot,
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
Thanks once again,
--j
--
http://mail.python.org/mailman/listinfo/python-list
John [EMAIL PROTECTED] writes:
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
Whoops, I meant to say:
len([1 for i in xrange(len(s1)) if s1.startswith(s2,i)])
That avoids creating a lot of small strings.
If s1 is large you may
John wrote:
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
Sure, to a point. Instead of:
def countoverlap(s1, s2):
return len([1 for i in xrange(len(s1)) if s1[i:].startswith(s2)])
Try this version, which takes smaller
John [EMAIL PROTECTED] wrote:
Thanks a lot,
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
Sure (untested code):
def count_with_overlaps(needle, haystack):
count = 0
pos = 0
while True:
where =
Alex Martelli [EMAIL PROTECTED] wrote:
John [EMAIL PROTECTED] wrote:
Thanks a lot,
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
Sure (untested code):
def count_with_overlaps(needle, haystack):
count = 0
